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A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300" rad s"^(-1) . (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. |
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Answer» Solution :(a) `I_("RMS") =V_("rms") omega C=6.9 mu A` (b) Yes. The derivation in Exercise 8.1(b) is true EVEN if i is oscillating in TIME. (c ) The FORMULA `B =(mu_(0) r)/(2 pi R^(2)) i_(d)` goes through even if `i_(d)` (and therefore B) oscillates in time. The formula shows they oscillate in phase. Since `i_(d)= i`, we have `B_(0)=(mu_(0) r)/(2 pi R^(2)) i_(0)`," where "B_(0) and i_(0)` are the amplitudes of the oscillating magnetic field and current, RESPECTIVELY. `i_(0) =sqrt(2) I_("rms")=9.76 mu A.` For `r=3 cm, R=6 cm, B_(0)=1.63xx10^(-11) T.` |
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